Inequality: 2a^2 + b^2 = 9c^2 implies 2c/a + c/b ≥ √3

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Inequality: 2a^2 + b^2 = 9c^2 implies 2c/a + c/b ≥ √3

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Source: Estonia, Team Selection Test for the IMO, 2011

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316 posts

Shu

#1 h Aug 5, 2011, 6:30 PM • 2 Y

Y by younesmath2012maroc, Adventure10

Let $a$ , $b$ , $c$ be positive real numbers satisfying $2a^2+b^2=9c^2$ . Prove that $\frac{2c}{a}+\frac{c}{b}\ge\sqrt{3}.$

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#2 h Aug 5, 2011, 6:36 PM • 2 Y

Y by Adventure10 and 1 other user

Hoelder is enough....

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#3 h Aug 6, 2011, 3:16 AM • 1 Y

Y by Adventure10

socrates wrote:

Hoelder is enough....

y dont we try trigonometry

$2a^2+b^2=9c^2$

there exists $t,0<t<\frac{\pi}{2}$ such that $sint=\frac{a\sqrt{2}}{3c}$ and $cost=\frac{b}{3c}$

$\frac{2c}{a}+\frac{c}{b}\geq\sqrt{3}.$
is equivalent to

$\frac{2\sqrt{2}}{3sint}+\frac{1}{3cost}\geq\sqrt{3}$

it's not hard from now. u can use $f'(t)$ to solve the problem

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arqady

#4 h Aug 6, 2011, 10:38 AM • 2 Y

Y by Adventure10, Mango247

creatorvn wrote:

$\frac{2\sqrt{2}}{3sint}+\frac{1}{3cost}\geq\sqrt{3}$

From here by Holder we obtain:
$\left(\frac{2\sqrt{2}}{3sint}+\frac{1}{3cost}\right)^2(\sin^2t+\cos^2t)\geq\left(\sqrt[3]{\frac{8}{9}}+\sqrt[3]{\frac{1}{9}}\right)^3=3$ .

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#5 h Aug 7, 2011, 9:22 AM • 2 Y

Y by Adventure10, Mango247

$\frac{2\sqrt{2}}{3sint}+\frac{1}{3cost}\geq\sqrt{3}$
$\frac{2\sqrt{2}}{3sint}+ \frac{3\sqrt{3}}{2}sin^{2}t\geq2[ \frac{3\sqrt{3}}{2}]$
$\frac{1}{3cost}+ \frac{3\sqrt{3}}{2}cos^{2}t\geq \frac{3\sqrt{3}}{2}$

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#6 h Aug 8, 2011, 7:09 AM • 1 Y

Y by Adventure10

by A_G inequality ：(sinx)^2+(cosx)^2=1，look my solution
$\frac{2\sqrt{2}}{3sint}+\frac{1}{3cost}\geq\sqrt{3}$
$\frac{2\sqrt{2}}{3sint}+\frac{3\sqrt{3}}{2}sin^{2}t\geq2[\frac{3\sqrt{3}}{2}]$

$\frac{1}{3cost}+\frac{3\sqrt{3}}{2}cos^{2}t\geq\frac{3\sqrt{3}}{2}$

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#7 h Aug 8, 2011, 6:50 PM • 4 Y

Y by ArrowRowe, Adventure10, Mango247, and 1 other user

Click to reveal hidden text

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#8 h Aug 9, 2011, 10:12 AM • 1 Y

Y by Adventure10

salution of kitsune = Holder

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sqing

#9 h Nov 30, 2012, 8:24 AM • 1 Y

Y by Adventure10

Equivalent to
For $a,b>0,a^2+b^2=9 ,$ prove that $\frac{2\sqrt{2}}{a}+\frac{1}{b}\ge\sqrt{3}\ .$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2719683

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younesmath2012maroc

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younesmath2012maroc

#10 h Nov 30, 2012, 10:52 AM • 3 Y

Y by sqing, Adventure10, Mango247

Shu wrote:

Let $a$ , $b$ , $c$ be positive real numbers satisfying $2a^2+b^2=9c^2$ . Prove that $\frac{2c}{a}+\frac{c}{b}\ge\sqrt{3}.$

$T=\frac{2c}{a}+\frac{c}{b}$ using holder we have $9T^2=(\frac{2}{a}+\frac{1}{b})(\frac{2}{a}+\frac{1}{b})(2a^2+b^2)\ge (2+1)^3$
thus $T \ge \sqrt{3}$

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#11 h Dec 10, 2012, 7:13 PM • 2 Y

Y by Adventure10, Mango247

$\frac{a^{2}}{c^{2}}+\frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=9\ge \frac{(\frac{a}{c}+\frac{a}{c}+\frac{b}{c})^2}{3}$
therefore
$(\frac{a}{c}+\frac{a}{c}+\frac{b}{c})\le 3\cdot \sqrt{3}$
we know
$(\frac{a}{c}+\frac{a}{c}+\frac{b}{c})(\frac{c}{a}+\frac{c}{a}+\frac{c}{b})\ge 9$
then $\frac{2c}{a}+\frac{c}{b}\ge\sqrt{3}$

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abdelkrim-amine

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abdelkrim-amine

#12 h Dec 10, 2012, 8:35 PM • 3 Y

Y by younesmath2012maroc, Adventure10, Mango247

Shu wrote:

Let $a$ , $b$ , $c$ be positive real numbers satisfying $2a^2+b^2=9c^2$ . Prove that $\frac{2c}{a}+\frac{c}{b}\ge\sqrt{3}.$

By AM-GM $2a^2+b^2 \geq 3\sqrt[3]{a^4b^2}$
So $\sqrt{3}c \geq \sqrt[3]{a^2b }\Leftrightarrow \frac{3c}{\sqrt[3]{a^2b}} \geq \sqrt{3}$
Finally $\frac{2c}{a} +\frac{c}{b} \geq \sqrt{3}$

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#13 h Mar 26, 2013, 6:49 PM • 1 Y

Y by Adventure10

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sqing

#14 h Jun 5, 2013, 2:12 AM • 2 Y

Y by Adventure10, Mango247

creatorvn wrote:

Click to reveal hidden text
is equivalent to

$\frac{2\sqrt{2}}{3sint}+\frac{1}{3cost}\geq\sqrt{3}$

it's not hard from now.
Click to reveal hidden text

$\dfrac{2\sqrt2}{sint}+\dfrac1{cost}=\dfrac4{\sqrt2 sint}+\dfrac1{cost}\geq\dfrac{9}{\sqrt2sint+cost}$ $=\dfrac{9}{\sqrt3sin(t+\alpha)}\ge\dfrac{9}{\sqrt3}=3\sqrt3.$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=3086954#p3086954

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sqing

#15 h Mar 27, 2015, 7:37 AM • 2 Y

Y by Adventure10, Mango247

Estonia 2004 ?:
Let $a,b,c>0, a^2+b^2+c^2=3 ,$ prove that $\frac{1}{1+2bc}+\frac{1}{1+2ca}+\frac{1}{1+2ab}\geq 1.$

This post has been edited 1 time. Last edited by sqing, Mar 27, 2015, 9:09 AM

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arqady

#16 h Mar 27, 2015, 8:16 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Estonia 2004 ?:
Let $a^2+b^2+c^2=3 ,$ prove that $\frac{1}{1+2bc}+\frac{1}{1+2ca}+\frac{1}{1+2ab}\geq 1.$

It's wrong of course! Try $1+2bc\rightarrow0^-$ .

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sqing

#17 h Mar 26, 2020, 2:01 AM • 3 Y

Y by Mango247, Mango247, Mango247

sqing wrote:

Estonia 2004 ?:
Let $a,b,c>0, a^2+b^2+c^2=3 ,$ prove that $\frac{1}{1+2bc}+\frac{1}{1+2ca}+\frac{1}{1+2ab}\geq 1.$

2004 Estonia National Olympiad
$\frac{1}{1+2bc}+\frac{1}{1+2ca}+\frac{1}{1+2ab}\geq \frac{9}{3+2(ab+bc+ca)}\geq \frac{9}{3+2(a^2+b^2+c^2)}=1.$

This post has been edited 1 time. Last edited by sqing, Mar 26, 2020, 3:03 PM

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